Final answer:
The energy released from condensation for a small storm with a 1 km radius and 1.0 cm of rain is approximately 7.0686 × 10^13 J, which corresponds to answer option (a) 1.08 × 10^14 J.
Step-by-step explanation:
The question involves calculating the energy released during condensation in a thunderstorm, specifically for rain precipitated over a certain area. The latent heat of condensation, which is the energy released when water condenses from vapor to liquid, is key to this calculation. We know from the information given that the latent heat of condensation for water is 2,250 J per gram.
First, we calculate the volume of rain using the formula for the volume of a cylinder (since a storm can roughly be approximated as a circular area) V = πr^2h, where r is the radius and h is the height or thickness of the rain layer. Then we convert this volume into mass by using the density of water (1 gram per cubic centimeter) and finally multiply by 2,250 J/g to find the total amount of energy released.
Calculation Details
For a storm of 1 km in radius with 1.0 cm of rain:
- Radius (r) = 1 km = 1000 meters = 100,000 cm
- Height (h) = 1.0 cm
- Volume (V) = π(100,000)^2(1) cm^3
- Volume (V) = 3.1416 × 10^10 cm^3
- Mass (m) = Volume (V) (density of water) = 3.1416 × 10^10 grams
- Energy released = Mass (m) × 2,250 J/g
- Energy released = 3.1416 × 10^10 g × 2,250 J/g
- Energy released = 7.0686 × 10^13 J approximately
Therefore, the closest answer from the provided options is (a) 1.08 × 10^14 J.